a ball projected vertically upwards from the top of the tower 60 meters high with a velocity of 30 meters

per second. what is the maximum height above the ground? how long does it take to reach the ground?

11 answers

the height is modeled by the function

h(t) = 60 + 30t - 4.9t^2

max height is at the vertex of the parabola, at t = -b/2a = 30/9.8

Use that to evaluate h(t) there.

It hits the ground when the height has dropped to zero. So, just solve

60 + 30t - 4.9t^2 = 0

V²=u²-2gh
At hmax v=0
0=30²-2x10xh
0=900-20h
20h=900
h= 900÷ 20
h=45
hmax = 45 + 60= 105 m

V²=u²-2gh
At hmax v=0
0=30²-2x10xh
0=900-20h
20h=900
h= 900÷ 20
h=45
hmax = 45 + 60= 105 m

For time
V=u + gt
0=30-10t
10t=30
t=30÷10
Time taken to reach the ground
h=ut+½gt²
U=0 when body falls downward
105=0xt +½x10xt²
105=5t²
t²= 21
t=4.5
Total time taken to reach the ground=>3+4.5=7.5 seconds.

Thanks

beauty i love you
your expression
really defined
your beauty

Thanks

Beauty i love u

I love the way you rote it

Baby thank u love 💕 you

For time
V=u + gt
0=30-10t
10t=30
t=30÷10
Time taken to reach the ground
h=ut+½gt²
U=0 when body falls downward
105=0xt +½x10xt²
105=5t²
t²= 21
t=4.5
Total time taken to reach the ground=>3+4.5=7.5 seconds.

The same as there's